The problem is as follows:
Alice and Willy had bought a new house. The figure from below is a sketch of the house. The couple wants the home to be surrounded completely by a wooden fence, which must occupy the maximum area as posible in front of the house. This is is indicated by a lighter shade in the drawing where $ABCD$ is a rectangle. The couple however, only got the materials including the wood, nails and paint from the hardware store to complete $98$ meters of the fence. Assuming the maximum area the couple wants to enclose starts in front of the house as indicated in the arrow up to the entrance of the house, How many meters in a straight line from the house will the entrance to the fence be located?
The alternatives in my book are as follows:
$\begin{array}{ll} 1.&18\,m\\ 2.&14\,m\\ 3.&16\,m\\ 4.&25\,m\\ \end{array}$
I'm assuming that to solve this problem it is required the use of derivatives, since it will be generated a quadratic equation. But I don't know how to establish such equation. Can someone help me with this part?.
Since what they require is to get the maximum area, I believe it will be given by:
$a(b+c+2)=A$
Assuming the small segments b and c are between the entrance door in the fence.
Since $a+b+c=98$
Then this means:
$a(98-a+2)=A(a)$
Then this would mean:
$A(a)=a(100-a)=100a-a^2$
Since the function will be as follows:
$A(a)=100a-a^2$
Thus the way to attain the maximum will be the derivate equal to zero.
$A'(a)=100-2a=0$
$a=50$
Therefore this would be the maximum.
But this answer does not appear in the alternatives. Which part did I made a mistake?. Can someone help me here?. Please an answer must include a drawing since for me it is difficult to spot where to establish the right equation in order to maximize it.
Your formula for the area is wrong.
Suppose that the side lengths AB and CD are each $x$ units.
Then, the width of the rectangle will be
$$98 - (2 \times x) + 2 = 100 - 2x.$$
This means that the area will be
$$f(x) = (100 - 2x) \times x = 100x - 2x^2.$$
Taking the derivative,
$$f'(x) = 100 - 4x.$$
This means that $f(x)$ will have a maximum when
$$100 - 4x = 0.$$
Addendum
Per OP's request:
Reactions to his recent questions.
First, see my comment that immediately follows your query.
I adopted the interpretation that the width of the fence is variable, only because that was the only way that the problem could be attacked.
Let's take it one step at a time.
There are two possibilities:
either the width of the fence is some fixed distance a.
or the width of the fence is variable.
If the width of the fence is fixed, at the value $a$, then the problem makes no sense! Naturally, you would then use all of the fence, so the length of each of the sides would be
$$\left(\frac{1}{2}\right) \times \left[98 - (a-2)\right] ~=~ \left[50 - \left(\frac{a}{2}\right)\right].$$
Since the assumption of a fixed-width fence leads to the conclusion that the dimensions of the fence are fixed, the assumption is untenable.
Assuming that the dimensions of the width of the fence equals the variable $a$, and that the dimensions of each side of the fence is $x$, then you have that
$$(a-2) + 2x = 98 \implies a = (100 - 2x).$$
Therefore, the problem is reduced to only dealing with one variable, $x$.
You have raised a separate question - why couldn't the problem be attacked using two variables, $a$ and $x$. Although this is feasible, as long as you remain aware that $a$ must equal $(100 - 2x)$ it adds an unnessary complication.
The main reason that you had trouble relates to the first sentence that I posted in my original answer: your formula for the area is wrong.
You had
$$a(b + c + 2) = A. \tag1$$
I soon as I saw that, I stopped reading your analysis, because that formula for the area makes no sense. If AB and CD have lengths $b$ and $c$ respectively, where $b = c$, then the area is
$$A = (a \times b).$$
If (for some bizarre reason), $b \neq c$, then you have a trapezoid and the area is
$$A = \left(a \times \frac{b+c}{2}\right).$$
Therefore, there was no way that your formula in line (1) above, was correct.
You raised another point:
I always assumed that points B and C were fixed. The variable $x$ was intended to represent the length of line BA, which I presumed to be equal to the length of line CD.
Another question
My original answer stands; I have already solved the problem. The key points in my solution are
The width of the fence is variable, rather than being the fixed length of line BC.
The problem can be reduced to a single variable problem (which is clearly best) by understanding that the width $a$ must equal $(100 - 2x).$